Question: Define a spherical shell centered at the origin: $S = \{ (x, y, z) \in \mathbb{R}^3 \big | 1 \leq x^2 + y^2 + z^2 \leq 6 \}$ What is the triple integral of $f(x, y, z) = y^3z$ over $S$ in spherical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_1^6 \rho^5\sin^3(\theta)\sin^3(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice B) B $ \int_0^\pi \int_0^{2\pi} \int_1^6 \rho^6\sin^3(\theta)\sin^4(\varphi)\cos(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_1^6 \rho^7\sin^3(\theta)\sin^2(\varphi)\cos(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice D) D $ \int_0^\pi \int_0^{2\pi} \int_1^6 \rho^4\sin^3(\theta)\sin^3(\varphi)\cos(\varphi) \, d\rho \, d\theta \, d\varphi$
Solution: The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the concentric spheres with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. Here, the region $S$ ranges from a radius of $1$ to a radius of $6$. Therefore, we want $1 < \rho < 6$. $ \int_0^\pi \int_0^{2\pi} \int_1^6 \cdots \, d\rho \, d\theta \, d\varphi$ We can now put $f(x, y, z)$ in the integrand, but we need to substitute $x$, $y$, and $z$ for their definitions in spherical coordinates. $ \int_0^\pi \int_0^{2\pi} \int_1^6 \rho^4\sin^3(\theta)\sin^3(\varphi)\cos(\varphi) \, d\rho \, d\theta \, d\varphi$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] The integral in spherical coordinates: $ \int_0^\pi \int_0^{2\pi} \int_1^6 \rho^6\sin^3(\theta)\sin^4(\varphi)\cos(\varphi) \, d\rho \, d\theta \, d\varphi$